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Question Number 121085 by TITA last updated on 05/Nov/20

Commented by TITA last updated on 05/Nov/20

$p l e a s e h e l p$
	Answered by Dwaipayan Shikari last updated on 05/Nov/20

	$3 x^{2} + x + 2 = 0 \Rightarrow x = \frac{- 1 \pm i \sqrt{23}}{6}$ $α + β = - \frac{1}{3} α β = \frac{2}{3}$ $\frac{α + β}{α β} = \frac{- 1}{2} \Rightarrow \frac{1}{α} + \frac{1}{β} = - \frac{1}{2} \Rightarrow \frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{2}{α β} = \frac{1}{4}$ $\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{1}{4} - 3 = - \frac{11}{4}$ $E q u a t i o n x^{2} - (\frac{1}{α^{2}} + \frac{1}{β^{2}}) x + {(\frac{1}{α β})}^{2} = x^{2} + \frac{11}{4} x + \frac{9}{4} = 0$ $4 x^{2} + 11 x + 9 = 0$
	Answered by liberty last updated on 05/Nov/20
	$(14) 3x^2 +x+2=0 → { (α),(β) :} → { ((α=((−1+(√(1−24)))/6) = ((−1+i(√(23)))/6))),((β=((−1−i(√(23)))/6))) :} → { ((α^2 =((−22−2i(√(23)))/(36))= ((−11−i(√(23)))/(18)))),((β^2 =((−11+i(√(23)))/(18)))) :} → { (((1/α^2 ) = ((−18)/(11+i(√(23)))) = ((−11+i(√(23)))/8))),(((1/β^2 )=((−18)/(11−i(√(23))))=((−11−i(√(23)) )/8))) :}$
	$(14) {3 x}^{2} + x + 2 = 0 \to {\begin{cases} α \\ β \end{cases}$ $\to {\begin{cases} α = \frac{- 1 + \sqrt{1 - 24}}{6} = \frac{- 1 + i \sqrt{23}}{6} \\ β = \frac{- 1 - i \sqrt{23}}{6} \end{cases}$ $\to {\begin{cases} α^{2} = \frac{- 22 - 2 i \sqrt{23}}{36} = \frac{- 11 - i \sqrt{23}}{18} \\ β^{2} = \frac{- 11 + i \sqrt{23}}{18} \end{cases}$ $\to {\begin{cases} \frac{1}{α^{2}} = \frac{- 18}{11 + i \sqrt{23}} = \frac{- 11 + i \sqrt{23}}{8} \\ \frac{1}{β^{2}} = \frac{- 18}{11 - i \sqrt{23}} = \frac{- 11 - i \sqrt{23}}{8} \end{cases}$
	Answered by TANMAY PANACEA last updated on 05/Nov/20
	$α+β=((−1)/3) αβ=(2/3) 3x^2 +x+2=0 x=((−1±(√(1−4×3×2)))/(2×3))=((−1±i(√(23)))/6) α=((−1+i(√(23)))/6) β=((−1−i(√(23)))/6) (1/α^2 )=((1−i×2(√(23)) −23)/(36))=((−22−i(√(23)))/(36)) (1/β^2 )=((1+i2(√(23)) −23)/(36))=((−22+i2(√(23)))/(36)) eqn x^2 −x((1/α^2 )+(1/β^2 ))+(1/((αβ)^2 ))=0 x^2 −x{(((α+β)^2 −2αβ)/((αβ)^2 ))}+(1/((αβ)^2 ))=0 x^2 −x((((1/9)−2×(2/3))/(4/9)))+(9/4)=0 x^2 −x(((1−12)/4))+(9/4)=0 4x^2 +11x+9=0 27α^4 =11α+10 α=((−1+i(√(23)))/6)→α^2 =((1−i2(√(23)) −23)/(36))=((−22−i2(√(23)))/(36))=((−11−i(√(23)))/(18)) α^4 =((121+i22(√(23)) −23)/(324))=((98+i22(√(23)))/(324))=((49+i11(√(23)))/(162)) 27α^4 =27×((49+i11(√(23)))/(162))=((49+i11(√(23)))/6) 11α+10 11×((−1+i(√(23)))/6)+10 ((−11+i11(√(23)) +60)/6) =((49+i11(√(23)))/6) so LHS=RHS$
	$α + β = \frac{- 1}{3}$ $α β = \frac{2}{3}$ $3 x^{2} + x + 2 = 0$ $x = \frac{- 1 \pm \sqrt{1 - 4 \times 3 \times 2}}{2 \times 3} = \frac{- 1 \pm i \sqrt{23}}{6}$ $α = \frac{- 1 + i \sqrt{23}}{6} β = \frac{- 1 - i \sqrt{23}}{6}$ $\frac{1}{α^{2}} = \frac{1 - i \times 2 \sqrt{23} - 23}{36} = \frac{- 22 - i \sqrt{23}}{36}$ $\frac{1}{β^{2}} = \frac{1 + i 2 \sqrt{23} - 23}{36} = \frac{- 22 + i 2 \sqrt{23}}{36}$ $e q n$ $x^{2} - x (\frac{1}{α^{2}} + \frac{1}{β^{2}}) + \frac{1}{{(α β)}^{2}} = 0$ $x^{2} - x {\frac{{(α + β)}^{2} - 2 α β}{{(α β)}^{2}}} + \frac{1}{{(α β)}^{2}} = 0$ $x^{2} - x (\frac{\frac{1}{9} - 2 \times \frac{2}{3}}{\frac{4}{9}}) + \frac{9}{4} = 0$ $x^{2} - x (\frac{1 - 12}{4}) + \frac{9}{4} = 0$ $4 x^{2} + 11 x + 9 = 0$ $27 α^{4} = 11 α + 10$ $α = \frac{- 1 + i \sqrt{23}}{6} \to α^{2} = \frac{1 - i 2 \sqrt{23} - 23}{36} = \frac{- 22 - i 2 \sqrt{23}}{36} = \frac{- 11 - i \sqrt{23}}{18}$ $α^{4} = \frac{121 + i 22 \sqrt{23} - 23}{324} = \frac{98 + i 22 \sqrt{23}}{324} = \frac{49 + i 11 \sqrt{23}}{162}$ $27 α^{4} = 27 \times \frac{49 + i 11 \sqrt{23}}{162} = \frac{49 + i 11 \sqrt{23}}{6}$ $11 α + 10$ $11 \times \frac{- 1 + i \sqrt{23}}{6} + 10$ $\frac{- 11 + i 11 \sqrt{23} + 60}{6}$ $= \frac{49 + i 11 \sqrt{23}}{6}$ $s o L H S = R H S$
	Commented by TITA last updated on 05/Nov/20

	$t h a n k s s i r$
	Commented by TANMAY PANACEA last updated on 05/Nov/20

	$m o s t w e l c o m e$
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